The solution would be like
this for this specific problem:
A = (l + 20) · (w + 10)
The area of the pool is 1800 square feet and can be computed
also like l · w. Then l · w = 1800, and therefore l = 1800 w.
We plug in the values:
A = (1800 / w + 20) · (w + 10) = 1800 + 18000 / w + 20w +
200 = 18000 / w + 20w + 2000
dA /dw = − 18000 / w2 + 20
w = 30
l = 60
Since A = (l + 20) · (w + 10):
A = (60 + 20) · (30+ 10)
A = 3,200
<span>A(x) = 1800 + 2*10(x+10)+ 2*5(1800/x) </span><span>
<span>= 1800 + 20x + 200 + 18000/x </span>
<span>= 2000 + 20 x + 18000/x </span></span>
<span>20 - 18000/x^2 = 0 </span><span>
<span>x = sqrt (900) = 30 </span>
<span>Pool length dimension = 1800/x = 60 </span>
Lot dimensions: 80 x 40</span>
I am hoping that these answers
have satisfied your queries and it will be able to help you in your endeavors, and
if you would like, feel free to ask another question.
Answer:
12.083
Explanation:
Plug into a calculator.
80% of their games are 28.8, or 29 games (since they need to win at least 80%). They have already won 16. 29-16 is 13.
Therefore, they need to win 13 more games.
P.S. The midseason statement is irrelevant.
Answer:
Step-by-step explanation:
This situation can be modeled by a binomial distribution of parameters:
We want to find the probability that at least 90 are in repair.
<u><em>We can approximate this problem to a normal distribution, where:</em></u>
Then we look for
Then we must find the normal standard statistic Z-score
Therefore:
Looking in the standard normal table we obtain:
Let's denote students in D = in the drama club , S<span> = in a sports team </span>
<span>P(D) = 85/330 </span>
<span>P(S) = 200/330 </span>
<span>
P(D and S) = 60/300 </span>
<span>
P(D or S) = P(D) + P(S) - P(D and S) </span><span>= 85/330 + 200/330 - 60/330 = 15/22 </span>
<span>
P(neither D nor S) = 1- 15/22 </span><span>= 7/22</span>
